Question

A plate of glass with parallel faces having a refractive index of 1.46 is resting on the surface of water in a tank. A ray of light coming from above in air makes anangle of incidence 35.0 with the normal to the top surface of the glass.
What angle does the ray refracted into the water make with the normal to the surface? Use 1.33 for the index of refraction of water.

Answer 1

Sin (Theta1) / Sin(Theta 2) = n_glass/n_air
n_glass = 1.46
n_air = 1
theta1 = 33.5o

Sin(33.5) / Sin(Theta2) = 1.46/1
1*Sin(33.5)/1.46 = Sin(theta2)
Theta2 = 22.21o


Theta2 = 22.21o
Theta3 = ??
n_glass = 1.46
n_water = 1.33

Sin(Theta2)/Sin(Theta3) = n_water/n_glass
Sin(22.21) / Sin(Theta3) = 1.33/1.46
Sin(Theta3) = 1.46/1.33 * sin(22.21)
Sin(Theta3) = 0.415
Theta3 = sin-1 (0.415) = 24.51 o

Answer 2

Given:Refractive index of glass μ=1.46Refractiv index of water μ = 1.33angle of incidence i = 35

According to snell's law of refraction of lightsin i/sinr = μ2/μ1sin35/sinr = 1.33/1.46sin35/sinr = 0.9109angle of refraction = sin^-1(0.6296)= 39.02 degree.

Hope that helps, let me know if you have any questions, good luck!

Answer 3

I think I see what I did wrong there,

sin θ1 / sinθ2 = ng / na ; ----------1
g : glass , a: air , w : water ,
sin θ2 / sin θ3 = nw / ng ; -----------2
1 multiplied by 2 gives ,
sin θ1 / sin θ3 = nw / na ;
sin 33.5 / sin θ3 = 1.33 / 1 ;
sin θ3 = sin 35/ 1.46 ;
or θ3 = 23.14 degrees, which is the answer.